# What is the equation of the line that passes through (-1,3) and is perpendicular to the line that passes through the following points: (- 2,4),(-7,2)?

Nov 12, 2017

See a solution process below:

#### Explanation:

First, we need to find the slope of the line which passes through $\left(- 2 , 4\right)$ and $\left(- 7 , 2\right)$. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$

Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.

Substituting the values from the points in the problem gives:

$m = \frac{\textcolor{red}{2} - \textcolor{b l u e}{4}}{\textcolor{red}{- 7} - \textcolor{b l u e}{- 2}} = \frac{\textcolor{red}{2} - \textcolor{b l u e}{4}}{\textcolor{red}{- 7} + \textcolor{b l u e}{2}} = \frac{- 2}{-} 5 = \frac{2}{5}$

A perpendicular slope is the negative inverse of the original slope. Let's call the perpendicular slope ${m}_{p}$.

The we can say: ${m}_{p} = - \frac{1}{m}$

Or, for this problem:

${m}_{p} = - \frac{1}{\frac{2}{5}} = - \frac{5}{2}$

We can now use the point-slope formula to find the equation of the line passing through $\left(- 1 , 3\right)$ with a slope of $- \frac{5}{2}$. The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.

Substituting the slope we calculated and the values from the point in the problem gives:

$\left(y - \textcolor{b l u e}{3}\right) = \textcolor{red}{- \frac{5}{2}} \left(x - \textcolor{b l u e}{- 1}\right)$

$\left(y - \textcolor{b l u e}{3}\right) = \textcolor{red}{- \frac{5}{2}} \left(x + \textcolor{b l u e}{1}\right)$

If we want this slope-intercept form we can solve for $y$ giving:

$y - \textcolor{b l u e}{3} = \left(\textcolor{red}{- \frac{5}{2}} \times x\right) + \left(\textcolor{red}{- \frac{5}{2}} \times \textcolor{b l u e}{1}\right)$

$y - \textcolor{b l u e}{3} = - \frac{5}{2} x - \frac{5}{2}$

$y - \textcolor{b l u e}{3} + 3 = - \frac{5}{2} x - \frac{5}{2} + 3$

$y - 0 = - \frac{5}{2} x - \frac{5}{2} + \left(\frac{2}{2} \times 3\right)$

$y = - \frac{5}{2} x - \frac{5}{2} + \frac{6}{2}$

$y = - \frac{5}{2} x + \frac{1}{2}$